coeforces 628C. Bear and String Distance

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题意：给一个字符串S，和 K 两个字符的dis == 其ASCII 码值得绝对值，

要求构造一个字符串S1是的sum = dis(S[i] - S1[i]) (i=0,,1 ,,,,,n ,) ==k ;

思路：贪心，使得前面以最大差出现，后面保持一致即可

#include<bits/stdc++.h>using namespace std;string s,s1;int  main(){    int n,m;    scanf("%d%d",&n,&m);    cin>>s;    for(int i =0; i<s.length(); i++)    {        if(m==0)        {            s1+= s[i];            continue;        }        int x = s[i]-'a';        int y = 'z' - s[i];        if(x>y)        {            if(m>=x)            {                s1 += 'a';                m-=x;            }            else            {                s1+= s[i]-m;                m = 0;            }        }        else        {            if(m>=y)            {                s1+='z';                m-=y;            }            else            {                s1+=s[i]+m;                m = 0;            }        }    }    if(m!=0)    {        cout<<"-1\n"<<endl;        return 0;    }    cout<<s1<<endl;}

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